In this my other question, Are the solutions of $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$ correct?, I show how to find the possible solutions for $x$ in the equation. $\sqrt{2}$ is one root and this is ok, but, what about $-\sqrt{2}$? Maybe it isn't a root, but how to strictly prove this? The prove given in these question (as we can read in the comments) looks wrong, so I'm searching for a correct one.

Calculation details

This was posted in the other question (as a answer, but the truth is just a huge comment, as I explained) and I think is pertinent to here too.

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Looks like $-\sqrt{2}$ isn't a solution for the equation, but I'm not sure. Looks like too, the power tower of a number should converge only on a specific interval ($[e^{−e},e^{1/e}]$).

But using Mathematica and the ProductLog function (wich the Lambert $W(z)$ function) we find some strange things:

-1.41421356237309504880168872421
-0.163093997943414854921937604558+0.59044 I
0.140921295793052749536215801866-0.044791 I
1.10008630700672531426983704055+0.50079 I
-0.268168781568546776692908102136-0.14235 I
0.894980750563013739735614892750-1.1090 I
-33.5835630157562847787187418023+29.118 I
6.49187847255812829134661655850*10^-46-1.5181*10^-45 I
1.00000000000000000000000000000+1.5134*10^-45 I
-1.41421356237309504880168872421-2.2930*10^-44 I
-0.163093997943414854921937604558+0.59044 I
0.140921295793052749536215801866-0.044791 I
1.10008630700672531426983704055+0.50079 I
-0.268168781568546776692908102136-0.14235 I
0.894980750563013739735614892750-1.1090 I

Taking the principal value of the exponentials ... 8 steps is very close to zero. Around $10^{-45}$. Then 9 steps is close to 1, and then at 10 starts over again. So if you are doing this numerically, you had better take many more than 45 places in your approximations!
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GEdgarDec 7 '11 at 22:35

Raising a negative number (or any complex number that is not a positive real) to a non-integer power is not a well-defined thing to do. The basic example is $(-1)^{1/2}$. Is that $i$ or $-i$? It just gets worse with irrational powers. $-\sqrt{2}^{-\sqrt{2}}$ could be interpreted as $\sqrt{2}\exp(-\pi i\sqrt{2})$, $\sqrt{2}\exp(-3\pi i\sqrt{2})$, $\sqrt{2}\exp(-5\pi i\sqrt{2})$, etc. and all of these are different. You might decide to enforce that all arguments always be reduced to values in $(-\pi,\pi]$, but that would be unnatural and give discontinuous operations.
–
alex.jordanDec 8 '11 at 5:02

3 Answers
3

This is the list of the absolute values of the iterates of $\small b=-\sqrt{2} $ . Because it has a very obvious 9-periodic pattern at the beginning I separated the trajectory in vertical columns of only 9 subsequent iterates.
(computed with Pari/GP, internal precision 1200 decimal digits)

It does look as if the eighth member isn't zero; a plot of the curves $\Re((-\sqrt 2)^{x+iy})=0$ and $\Im((-\sqrt 2)^{x+iy})=0$ in the vicinity of the seventh member of the cycle gives a strong hint:

(An analytical proof of this will still be needed, though.)

Here are plots of the complex plane trajectory. Since the seventh member of the cycle is a bit larger in magnitude than the others, I have also thrown in a close-up:

This paper by Galidakis seems to give some conditions for a power tower iteration (see page 990 for instance), but I seem to have some difficulty applying his criterion for determining the cycle length of the power tower iteration. It involves a multivariate generalization of the Lambert function which Mathematica doesn't have, so I'll need to study this a bit more and see if Galidakis's results can be adapted for this situation.

Hmm, concerning the exactness of fp at 0 and 1: no, the small deviation from the integers remains. Consider the 9-fold iteration by the function $\small x_{k+1}=it9(x_k) $ then the first-order difference d1 between $\small x_0=1 \qquad x_1=it9(x_0) \qquad d_1=x_1-x_0 $ is in the near of 1e-40 and from this the next iteration moves only by about 1e-80 and from this only about 1e-120 and so on. That quality of approximation is suprisingly nearly geometrical; the quotient between subsequent differences goes quickly to a constant near a value of again something like 1e-40. (...cont...)
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Gottfried HelmsDec 10 '11 at 9:09

(...cont...) so possibly (this is only a speculation!) one can implement an acceleration of approximation when that geometric rate (with approximate quotient q) could somehow be inserted by the closed-form of the geometric series using (1-q) in the denominator of some improved formula...
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Gottfried HelmsDec 10 '11 at 9:12

This is something curious. Define the function $\small it9(x) = for(k=1,9,x=\exp(bl*x));return(x) \qquad \text{ where } bl=\log(-sqrt(2)) $. Then, using Pari/GP, one can ask for the power series of $\small it9(x-1) $ . We get $ \displaystyle it9(x-1) = (0.140921295793 - 0.0447908983417*I) + (0.E-1821552 + 0.E-1821552*I)*x + \ldots $ where all following coefficients are machine-epsilon. Surprisingly the constant is the 4'th or 5'th iterate of $\small x_{k+1}=exp(bl*x_{k})$ beginning at $\small x_0=0 $. Perhaps this helps for focusing the analytical reasoning?
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Gottfried HelmsDec 10 '11 at 22:05